Looking at the datasheet of the 78XX voltage regulators re 'power dissipation', it is clear that dropping from 12V to 5V - and drawing 1A is too much.

i have a 12V DC adaptor, can i then use a 7809 *and* a 7805 to drop the voltage over two 'regulators' - ie. dropping from 12V to 9V and then subsequently to 5V.

lousy efficiency aside, is this setup workable ?

i just want to test a circuit at minimal requirement of 1A but the only wall-wart i have that can supply 1000mA is a 12V one.

i don't intend to make it a fixed/permanent scenario.

on a related issue, if i were to purchase a DC-DC buck converter (lots of them on eBay based on the LM2596) - what would the 'current' situation be, meaning, are buck converters able to supply the full current coming in from the high voltage side ?

ie. 12V in at 1A, will come out at the lower voltage, but also at 1A.

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7805 will handle 1A at 12V input fine, so long as you use a big enough heatsink.

You'll need heatsinks anyway as 7W is too much for two bare TO220 packages. 1W perpackage (without heatsink) is pushing it, use heatsink!

But you'll be better off using a cheap switch-mode DC-DC converter like the eBay LM2596 units,much higher efficiency, no heatsink needed (only downside is the noise which could be an issuewith sensitive low-voltage analog signals).

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But you'll be better off using a cheap switch-mode DC-DC converter like the eBay LM2596 units,much higher efficiency, no heatsink needed (only downside is the noise which could be an issuewith sensitive low-voltage analog signals).

okay, looks like i'll be going with that option - how about the current, those converters will pass the input supply current through with no losses ?

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i have a 12V DC adaptor, can i then use a 7809 *and* a 7805 to drop the voltage over two 'regulators' - ie. dropping from 12V to 9V and then subsequently to 5V.

You could put 2 7805's parallel, guess you still need a heatsink, though a fairsize piece of Alu would do the trick the 7805's will tell you if it's okay, they have automatic shut down for over-current and over-heat.

Okay, looks like i'll be going with that option - how about the current, those converters will pass the input supply current through with no losses ?

It doesn't work like that. The short answer is don't worry, it will do what you need.

A DC DC converter swaps current for voltage. If you feed 12V in and take 5V out at 1A then the input current at 12V will be a lot less. If there were no losses then the input current at 12V for 5V 1A out would be 5/12 * 1 = 0.42A (This is where some kind person tells me I got the maths wrong!) However, they are not perfect, they do waste some power so I would assume that for 1A out at 5V then 0.5A in at 12V would be needed.

A DC DC converter swaps current for voltage. If you feed 12V in and take 5V out at 1A then the input current at 12V will be a lot less. If there were no losses then the input current at 12V for 5V 1A out would be 5/12 * 1 = 0.42A (This is where some kind person tells me I got the maths wrong!)

Just to be clear a 7805 is not a converter like Perry is suggesting but a regulator, which drops voltage leaving current the same and dissipating the extra energy as heat, not very energy-efficient of course.

I see - I kinda figured that the power (wattage) is the limiting factor.

Conversely, with a boost converter, we can then expect a *drop* in the current ?(eg. from a 5V@1A supply, we won't get 1A if we boost it to 9V)

Okay - i shall proceed with confidence !

Power is a limiting factor, read the data sheet for the DC DC converter. However, for what you have described, I don't think you will have a problem as long as the converter you are using is rated to supply at least 1A. If a DC DC converter were 100% efficient (obviously they are not) then the input power would equal the output power, so Vin * Iin would equal Vout * Iout.

Yes, if you use a boost converter then the output current will be lower than the input current.

i see, looks like i still can't do the test yet then, no heatsinks yet.

just to be sure though, from the datasheet;it says Maximum Power Dissipation is PD=(TJ(max)- TA) / *0JA (*0 is actually theta)that becomes PD=(150-33)/19 = 6.16Wso; it won't handle 7W, right ?

That's using a PCB as heatsink or nothing as heatsink, using a block of copper that 19 becomes 3, and the power handling is 30W fora 90 deg rise above ambient. Personally I'd not recommend trying to more than 20W out of a TO220 packageas you have to allow for the heatsink itself and any thermal pad or grease.

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Perhaps cut up a couple of aluminium cans, open them out and arrange three or four layers together with a hole through them, use a washer to hold them together where you screw them to the tab of the 7805. The separate layers can fan out as fins on the heatsink.

Just to be clear a 7805 is not a converter like Perry is suggesting but a regulator, which drops voltage leaving current the same and dissipating the extra energy as heat, not very energy-efficient of course.

yes, thanks for confirming this - it's what i had in mind with my rudimentary knowledge of electronics.

That's using a PCB as heatsink or nothing as heatsink, using a block of copper that 19 becomes 3, and the power handling is 30W fora 90 deg rise above ambient....

ahh, i see - that's very handy to know - the datasheet wasn't very informative on that, it didn't even mention heatsinks at all which was strange - although i can't say what's normal, i think i've seen *a* datasheet which gave specs with heatsink usage advice but i don't know if that's the norm.

Perhaps cut up a couple of aluminium cans, open them out and arrange three or four layers together with a hole through them, use a washer to hold them together where you screw them to the tab of the 7805. The separate layers can fan out as fins on the heatsink.

No, I meant a USB phone charger that plugs into the cigarette lighter hole in the dashboard of a car.You just asked for a 12>5volt/1A regulator.A 5volt/1A phone/tablet USB supply is ofcourse an easy way to get 5volt/1A.Leo..